Question: Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}5x+8y &= 3 \\ -4x-4y &= -1\end{align*}$
Begin by moving the $y$ -term in the second equation to the right side of the equation. $-4x = 4y-1$ Divide both sides by $-4$ to isolate $x$ $x = {-y + \dfrac{1}{4}}$ Substitute this expression for $x$ in the first equation. $5({-y + \dfrac{1}{4}}) + 8y = 3$ $-5y + \dfrac{5}{4} + 8y = 3$ Simplify by combining terms, then solve for $y$ $3y + \dfrac{5}{4} = 3$ $3y = \dfrac{7}{4}$ $y = \dfrac{7}{12}$ Substitute $\dfrac{7}{12}$ for $y$ in the top equation. $5x+8( \dfrac{7}{12}) = 3$ $5x+\dfrac{14}{3} = 3$ $5x = -\dfrac{5}{3}$ $x = -\dfrac{1}{3}$ The solution is $\enspace x = -\dfrac{1}{3}, \enspace y = \dfrac{7}{12}$.